What is the integral of cot^2(x)csc^6(x) ?

The integral of cot^2(x)csc^6(x) is -1/7 cot^7(x)-2/5 cot^5(x)-1/3 cot^3(x)+C

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integral of cot^2(x)csc^6(x)

\int cot^{2} (x) csc^{6} (x) d x

- \frac{1}{7} cot^{7} (x) - \frac{2}{5} cot^{5} (x) - \frac{1}{3} cot^{3} (x) + C

Solution steps

\int cot^{2} (x) csc^{6} (x) d x

Apply Trigonometric Substitution

: \int - u^{2} (u^{2} + 1)^{2} d u

= \int - u^{2} (u^{2} + 1)^{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int u^{2} (u^{2} + 1)^{2} d u

Expand

u^{2} (u^{2} + 1)^{2} : u^{6} + 2 u^{4} + u^{2}

= - \int u^{6} + 2 u^{4} + u^{2} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (\int u^{6} d u + \int 2 u^{4} d u + \int u^{2} d u)

\int u^{6} d u = \frac{u ^{7}}{7}

\int 2 u^{4} d u = \frac{2 u ^{5}}{5}

\int u^{2} d u = \frac{u ^{3}}{3}

= - (\frac{u ^{7}}{7} + \frac{2 u ^{5}}{5} + \frac{u ^{3}}{3})

Substitute back

u = cot (x)

= - (\frac{cot ^{7} ( x )}{7} + \frac{2 cot ^{5} ( x )}{5} + \frac{cot ^{3} ( x )}{3})

Simplify

- (\frac{cot ^{7} ( x )}{7} + \frac{2 cot ^{5} ( x )}{5} + \frac{cot ^{3} ( x )}{3}) : - \frac{1}{7} cot^{7} (x) - \frac{2}{5} cot^{5} (x) - \frac{1}{3} cot^{3} (x)

= - \frac{1}{7} cot^{7} (x) - \frac{2}{5} cot^{5} (x) - \frac{1}{3} cot^{3} (x)

Add a constant to the solution

= - \frac{1}{7} cot^{7} (x) - \frac{2}{5} cot^{5} (x) - \frac{1}{3} cot^{3} (x) + C

\frac{x ^{2.3} + 4}{x ^{3.5} + 1}

\int w ln (w) d w

y^{^{'}} + \frac{3}{2 x} y = 0

\int_{1}^{5} (4 x - 1) d x

\int ln (x y) d x

What is the integral of cot^2(x)csc^6(x) ?
The integral of cot^2(x)csc^6(x) is -1/7 cot^7(x)-2/5 cot^5(x)-1/3 cot^3(x)+C